Question

What is the arclength of `(t^2lnt,(lnt)^2)` on `t in [1,2]`?

Answer 1

I used Wolframalpha to evaluate the integral.

`L = 2.815`

Explanation:

Here is a reference for Arc length of parametric equations

From the reference, the equation for arclength is:

`L = int_alpha^beta sqrt((dx/dt)^2 + (dy/dt)^2)dt`

`dx/dt = (d[t^2ln(t)])/dt = 2tln(t) + t`

`dy/dt = (d[(ln(t))^2])/dt = (2ln(t))/t`

I used Wolframalpha to evaluate the integral:

`L = int_1^2 sqrt((2tln(t) + t)^2 + ((2ln(t))/t)^2)dt = 2.815`